Difference between revisions of "1953 AHSME Problems/Problem 8"
Katzrockso (talk | contribs) (Created page with "==Problem 8== The value of <math>x</math> at the intersection of <math>y=\frac{8}{x^2+4}</math> and <math>x+y=2</math> is: <math>\textbf{(A)}\ -2+\sqrt{5} \qquad \textbf{(B...") |
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== Solution == | == Solution == | ||
<math>x+y=2\implies y=2-x</math>. Then <math>2-x=\frac{8}{x^2+4}\implies (2-x)(x^2+4)=8</math>. We now notice that <math>x=0\implies (2)(4)=8</math>, so <math>\fbox{C}</math> | <math>x+y=2\implies y=2-x</math>. Then <math>2-x=\frac{8}{x^2+4}\implies (2-x)(x^2+4)=8</math>. We now notice that <math>x=0\implies (2)(4)=8</math>, so <math>\fbox{C}</math> | ||
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+ | ==See Also== | ||
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+ | {{AHSME 50p box|year=1953|num-b=7|num-a=9}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 20:40, 1 April 2017
Problem 8
The value of at the intersection of and is:
Solution
. Then . We now notice that , so
See Also
1953 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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